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Linear algebra - operators

  1. Jul 31, 2008 #1
    1. The problem statement, all variables and given/known data
    I have a differentiation operator on P_3, and:

    S = {p \in P_3 | p(0) = 0}.

    I have to show that

    1) D : P_3 -> P_2 is not one-to-one.

    2) D: S -> P_3 is one-to-one.

    3) D: S -> P_3 is not onto.

    3. The attempt at a solution

    For #1, I want to show that our differentiation operator is not one-to-one by looking at the matrix that represents P_3 relative to P_2:

    [tex]A = \left( {\begin{array}{*{20}c}
    2 \hfill & 0 \hfill & 0 \hfill \\
    0 \hfill & 1 \hfill & 0 \hfill \\
    \end{array}} \right)[/tex]

    From this matrix, how do I show this?

    2 + 3) In this case, S is [0; bx; ax^2]. Again, I will use the same approach as in #1, but what is the connection between the matrix A and the operator being one-to-one or onto?
  2. jcsd
  3. Jul 31, 2008 #2


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    A function is "one-to-one" if and only if f(x)= f(y) means x= y. That's probably the simplest way to do this problem. Any function in P3 is of the form c+ ax+ bx2 as you say. Can you find two such polynomials that have the same derivative?

    If you really want to use matrices, Ax= Ay if and only if x= y is again the condition for being "one-to-one" so if A is not "one-to-one" the there exist distinct x and y such that Ax= Ay. That is the same as saying Ax- Ay= A(x- y)= 0 with x- y not equal to 0.
    Look for a vector z, in R3, such that Az= 0. If you take z= (u, v, w), what equations does Az= (0,0) give (and now you are back to my first suggestion).

    For problem (2), you are only looking, as you say, to polynomials of the form ax+ bx2. What is the difference between this and problem (1)?

    Can you see that it is NOT "onto" because the derivative of a quadratic is always linear- i.e. does not have an x2.

    I would not recomend trying to use matrices for this- just use the definition of D.
  4. Jul 31, 2008 #3
    Ok, I found what I was looking for in my book.

    It says: "A linear transformation f is one-to-one if the dimension of the kernel of f is zero" and "A linear transformation f : R^n -> R^m is onto if m =< n, that is, if the dimension of the image is equal or greater than n.".

    Can you confirm this? It seems to be a very easy way to solve my problem.
  5. Jul 31, 2008 #4


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    The first is just what I said before: one-to-one means that Ax= Ay only if x= y. Ax= Ay gives Ax- Ay= A(x-y)= 0. That is, a linear transformation is one-to-one only if Av= 0 means v= 0: that the kernel is only the 0 vector.

    Surely your book does not say "m<= n" and "the dimension of the image is equal or greater than n". Those two statements are contradictory. If f is defined on Rn, then its image can't be greater than n. A linear transformation is onto if its image has the same dimension as the domain space, here n.
  6. Jul 31, 2008 #5
    You are absolutely right - sorry.

    Thanks for your help.
  7. Jul 31, 2008 #6
    I'm trying to solve #3 and #4 using our discussion.

    #3: The matrix A that represents our operator D with respect to S and P_3 given by:

    [tex]A = \left( {\begin{array}{*{20}c}
    0 & 1 & 0 \\
    0 & 0 & 2 \\
    0 & 0 & 0 \\
    \end{array}} \right)[/tex]

    The nullspace of this matrix is spanned by e_1? So it is not one-to-one?

    #4: Using the matrix A, I find that the dimension of the image equals two, which is different from 3 (the dimension of S). So it is not onto - qed.
  8. Jul 31, 2008 #7
    My old heuristic way of thinking about it was to think of it this way: if a function is onto, you can imagine that every element in the image has some element of the domain mapping to it, right? So you have to "use" all of the elements in the image. If there were more objects in the image than in the domain, how could you construct a function that would "use" all the objects? You couldn't without violating the definition of a function.

    So for the linear operator to be onto, the dimension of the image better be less than or equal to the dimension of the vector space you're mapping from (and equal to the dimension of the function's domain).

    That being said, I think that the easiest way to show 1-1 is just to show that the dimension of the nullspace (or kernel) is zero. And obviously the only way you can have a linear operator be both 1-1 and onto is for the function to be between vector spaces of the same dimension.

    Sorry if this was information you already knew. Linear algebra is just so much fun :rofl: I couldn't help myself
  9. Aug 1, 2008 #8
    Hmm, I need to get some things cleared up. E.g. when we have a linear operator L : R^3 -> R^2, then R^3 is the domain and R^2 is the range, correct?

    Is the domain equal to the vector space I'm mapping from?

    I agree. But this is what I can't get - am I missing something here?
  10. Aug 1, 2008 #9


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    Yes, the domain is the vector space you are mapping from. The image is a subspace of the space you are mapping to since a linear transformation is not necessarily "onto".
  11. Aug 1, 2008 #10


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    R^2 is the codomain. The range (or image) of a function (or operator) L is the set {L(x) in R^2 | x in R^3}. If the range equals the codomain, then the function is onto.

  12. Aug 1, 2008 #11
    And does these terms also work for a function? So if we have a function f : A -> R^2 (A is the domain), then R^2 is not the range (image) of the function, but only the co-domain? Because in my Calculus-book, the author does not distinguish between these two terms.
    Last edited: Aug 1, 2008
  13. Aug 1, 2008 #12


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    Unfortunately, the term "range" is not always used precisely. If we have the function f:R-> R defined by f(x)= x2, some texts will say that the "range" is R, while others will say it is only the non-negative real numbers. Better to use "co-domain" and "image" which are better defined: the "co-domain" of this function is R, just because I said "R-> R" and the "mage" is the non-negative real numbers.
  14. Aug 1, 2008 #13


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    Of course, since a linear operator is also a function (a mapping).
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